Question

A hospital glucose solution is analyzed to verify its concentration. A \(10.0 \mathrm{~mL}\) sample with a mass of \(10.483 \mathrm{~g}\) is evaporated to dryness. If the solid glucose residue has a mass of \(0.524 \mathrm{~g},\) what are \((\mathrm{a})\) the mass \(/ \mathrm{mass}\) percent concentration, and (b) the molar concentration of the glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) solution?

Short answer

(a) 5.00% and (b) 0.291 M.

Step-by-step solution

Calculate Mass/Mass Percent

The mass/mass percent concentration is calculated using the formula:\[\text{Mass/Mass Percent} = \left( \frac{\text{Mass of solute}}{\text{Mass of solution}} \right) \times 100 \%\]Given, mass of the solute (glucose) is \(0.524 \text{ g}\) and mass of the solution is \(10.483 \text{ g}\).Substituting the values:\[\text{Mass/Mass Percent} = \left( \frac{0.524}{10.483} \right) \times 100 \% \approx 5.00 \%\]

Convert Mass to Moles

To find the molar concentration, we first need to convert the mass of glucose to moles. The molar mass of glucose \( (\text{C}_{6}\text{H}_{12}\text{O}_{6}) \) is calculated as follows:\[\text{Molar mass of } \text{C}_{6}\text{H}_{12}\text{O}_{6} = (6 \times 12.01) + (12 \times 1.01) + (6 \times 16.00) = 180.18 \text{ g/mol}\]Using the molar mass, convert the mass of glucose to moles:\[\text{Moles of glucose} = \frac{0.524 \text{ g}}{180.18 \text{ g/mol}} \approx 0.00291 \text{ moles}\]

Calculate Molar Concentration

Molar concentration is calculated using the formula: \[\text{Molarity} (M) = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\]The volume of the solution is given as \(10.0 \mathrm{~mL} = 0.0100 \text{ L}.\)Substituting the values:\[\text{Molarity} = \frac{0.00291 \text{ moles}}{0.0100 \text{ L}} = 0.291 \text{ M}\]

Key concepts

Mass/Mass Percent

Mass/mass percent is a way to express how much of a solute is present in a given amount of solution. It is calculated using the ratio of the mass of the solute (the substance being dissolved) to the total mass of the solution. This ratio is then multiplied by 100 to give a percentage.
In this exercise, the glucose solution's mass/mass percent is computed by taking the mass of glucose (0.524 g) and dividing it by the mass of the entire sample solution (10.483 g). Multiplying the resulting fraction by 100 yields:

• Formula: Mass/Mass Percent = \(\left( \frac{\text{Mass of solute}}{\text{Mass of solution}} \right) \times 100\%\)
• Calculation: \(\left( \frac{0.524 \text{ g}}{10.483 \text{ g}} \right) \times 100\% = 5.00\%\)

This result means that 5.00% of the solution's mass is due to glucose. Understanding this concept is crucial in fields like medicine, where precise concentrations impact treatment effectiveness.

Molarity

Molarity is used to express the concentration of a solution in terms of moles of the solute per liter of solution. This measure helps understand how many molecules or ions of a solute are present in a given volume of solution.
To find the molarity of a glucose solution in this exercise, we first need to calculate the moles of glucose from its given mass. Using the molar mass of glucose (180.18 g/mol), convert the mass of glucose to moles:

• Formula: Moles = \(\frac{\text{mass of glucose}}{\text{molar mass of glucose}}\)
• Calculation: \(\frac{0.524 \text{ g}}{180.18 \text{ g/mol}} \approx 0.00291 \text{ moles}\)

Next, use the moles of glucose and the volume of the solution in liters (0.0100 L) to calculate molarity:

• Formula: Molarity \(M = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\)
• Calculation: \(\frac{0.00291 \text{ moles}}{0.0100 \text{ L}} = 0.291 \text{ M}\)

This 0.291 M (molar) glucose solution is indicative of the number of glucose molecules dissolved per liter of solution, an essential computation in laboratory settings.

Glucose Solution

A glucose solution involves dissolving glucose, a simple sugar, into a solvent like water to form a homogeneous solution. This type of solution is frequently used in medical and scientific applications because glucose is a vital energy source for the body's cells.
When creating a glucose solution, concentration and purity are vital, especially in a medical context. The concentration determines the solution’s strength and efficacy for its intended purpose, such as intravenous solutions for patients requiring glucose. In this exercise, we understand how the concentration of glucose in a sample is computed and expressed both as mass/mass percent and molarity:

• Mass/mass percent reflects glucose as a proportion of the entire solution mass.
• Molarity shows how many moles of glucose exist per liter of solution, essential for precise dosage calculations.

Understanding the preparation and analysis of glucose solutions helps ensure their proper use in health care and laboratory processes.